Related Formula
Using Cramer's Rule:
If D = 0$D = 0$ and at least one of D₁, D₂, D₃$D_1, D_2, D_3$ is non-zero, the system has no solution.
If D = 0$D = 0$ and D₁ = D₂ = D₃ = 0$D_1 = D_2 = D_3 = 0$, the system generally has infinitely many solutions.
Core Logic
First, evaluate the main determinant D$D$:
D = vmatrix 1 & 1 & 1 2 & 5 & a 1 & 2 & 3 vmatrix$$D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{vmatrix}$$
D = 1(15 - 2a) - 1(6 - a) + 1(4 - 5) = 15 - 2a - 6 + a - 1 = 8 - a$$D = 1(15 - 2a) - 1(6 - a) + 1(4 - 5) = 15 - 2a - 6 + a - 1 = 8 - a$$
Setting D = 0 a = 8$D = 0 \implies a = 8$.
Now, evaluate D₃$D_3$ to check the consistency condition:
D₃ = vmatrix 1 & 1 & 6 2 & 5 & 36 1 & 2 & b vmatrix$$D_3 = \begin{vmatrix} 1 & 1 & 6 \\ 2 & 5 & 36 \\ 1 & 2 & b \end{vmatrix}$$
D₃ = 1(5b - 72) - 1(2b - 36) + 6(4 - 5) = 5b - 72 - 2b + 36 - 6 = 3b - 42$$D_3 = 1(5b - 72) - 1(2b - 36) + 6(4 - 5) = 5b - 72 - 2b + 36 - 6 = 3b - 42$$
Setting D₃ = 0 3b - 42 = 0 b = 14$D_3 = 0 \implies 3b - 42 = 0 \implies b = 14$.
Step 1: Confirming Infinitely Many Solutions
For a=8$a=8$ and b=14$b=14$, evaluate D₁$D_1$ and D₂$D_2$ to ensure they are also zero.
D₁ = vmatrix 6 & 1 & 1 36 & 5 & 8 14 & 2 & 3 vmatrix = 0$$D_1 = \begin{vmatrix} 6 & 1 & 1 \\ 36 & 5 & 8 \\ 14 & 2 & 3 \end{vmatrix} = 0$$
D₂ = vmatrix 1 & 6 & 1 2 & 36 & 8 1 & 14 & 3 vmatrix = 0$$D_2 = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 36 & 8 \\ 1 & 14 & 3 \end{vmatrix} = 0$$
Since D = D₁ = D₂ = D₃ = 0$D = D_1 = D_2 = D_3 = 0$, the system possesses infinitely many solutions.
Pattern Recognition
For a 3 × 3$3 \times 3$ system heavily relying on constants, D=0$D=0$ determines the parameter 'a' acting on the variable array, and D₃=0$D_3=0$ immediately forces the final constant 'b'.
Chapter Mix
Class 12 Maths: Matrices and Determinants